Thread: Help with url

Hi. Hope someone can give me a clue.

I have a java standalone using Eclipse EE. I have an embedded browser which works fine if I throw 'bbc.com' at it.

I have a text box where the users enters the url. A common entry would be '127.0.0.1:6500'.

However using URI and URL having no luck and just don't see it. Tried lots of combinations. Any hints very welcome.

Here my Listener.

buttonurl.addSelectionListener(new SelectionAdapter() {
        	@Override
            public void widgetSelected(SelectionEvent e) {
        		URI uri = null;
        		URL url = null;
        		try {
					uri = new URI(texturl.getText());
				} catch (URISyntaxException e1) {
					System.out.println("failed setting uri");
	        		e1.printStackTrace();
				}
        		System.out.println(uri.toString());
        		try {
					url = uri.toURL();
 
				} catch (MalformedURLException e1) {
					System.out.println("failed creating url");
	        		e1.printStackTrace();
				}
        		browser.setUrl(url.toString());
		}
        });

This code even with simple bbc.com gives me the stack trace below. But if I pass bbc.com as a simple string no problem,

Exception in thread "main" java.lang.IllegalArgumentException: URI is not absolute
at java.net.URI.toURL(Unknown Source)
at Path01class$31.widgetSelected(Path01class.java:156 6)
at org.eclipse.swt.widgets.TypedListener.handleEvent( TypedListener.java:248)
at org.eclipse.swt.widgets.EventTable.sendEvent(Event Table.java:84)
at org.eclipse.swt.widgets.Display.sendEvent(Display. java:4353)
at org.eclipse.swt.widgets.Widget.sendEvent(Widget.ja va:1061)
at org.eclipse.swt.widgets.Display.runDeferredEvents( Display.java:4172)
at org.eclipse.swt.widgets.Display.readAndDispatch(Di splay.java:3761)
at Path01class.main(Path01class.java:2352)
bbc.com

And if I try 127.0.0.1:6500 I get

java.net.URISyntaxException: Illegal character in scheme name at index 0: 127.0.0.1:6500
failed setting uri

having no luck

Please copy the full text of the error messages and paste it here.

Read the API doc for the URL class. Also it has a link to a site that defines what goes in a valid URL.
The String you show: '127.0.0.1:6500' is missing the protocol.